Ask your own question, for FREE!
Mathematics 23 Online
OpenStudy (anonymous):

You roll a standard number cube. Find P(number greater than 1)

OpenStudy (anonymous):

Chance of it being a 1?

OpenStudy (anonymous):

i think so lol

OpenStudy (anonymous):

it gives answers to choose from a)6/5 b)5/6 c)1/6 d)1

OpenStudy (anonymous):

Heh, P(not 1) is 1-P(1)

OpenStudy (anonymous):

i'm confused lol

OpenStudy (anonymous):

Do you know what the probability of rolling a 1 is?

OpenStudy (anonymous):

1/6 ?

OpenStudy (anonymous):

Right, so what you want is P(not 1), right?

OpenStudy (anonymous):

right

OpenStudy (anonymous):

a number greater then 1 though

OpenStudy (anonymous):

That's the same as (not 1)

OpenStudy (anonymous):

oh lol

OpenStudy (anonymous):

The probability of an event ocurring and not ocurring must sum to 1

OpenStudy (anonymous):

still confused lol

OpenStudy (anonymous):

Why is the probability of rolling a 1 equal to 1/6?

OpenStudy (anonymous):

because a normal cube goes up to 6

OpenStudy (anonymous):

Right, there are 6 possible outcomes and you want only one of them ->1/6 So the total probability is 6/6 ie 1 So if you are not getting a 1, you are getting any one of the others, what must be the probability of that?

OpenStudy (anonymous):

1/6?

OpenStudy (anonymous):

That is the probability of getting a 1 (or a 2, or a 3 etc), a specific number. What we want is the probability of it being ANYTHING other than 1.

OpenStudy (anonymous):

How many possibilities is that?

OpenStudy (anonymous):

5/6 ?

OpenStudy (anonymous):

Bingo! 1/6 + 5/6 = 1 P(event) + P(not event) = 1

OpenStudy (anonymous):

thanks (:

OpenStudy (anonymous):

ur welcome

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!